Re: Potentials, real and unreal

On Oct 8, 7:05 am, Bruce Scott TOK <Use-Author-Supplied-Address-
Header@[127.1]> wrote:

Ed Green wrote:
It happens that the divergence free part of the stress tensor has a
definite physical meaning. A divergence free stress field is a static
stress field (one not accelerating the medium), and hydrostatic
pressure is one example of such a field. Since different hydrostatic
pressures have different physical consequences, there is clearly
something "real" in this part of the description.

Only if there is a gradient (hence the usefulness of thinking of p as
part of a potential).

Reading that back, you seem to be saying that if there is no gradient
(of what?) then either (1) different hydrostatic pressures do not have
different physical physical consequences, or (2) it doesn't follow
that there is something real about the divergence free part of the
stress tensor.

Neither possibility really makes sense to me, so I'm not sure what you
are saying.

BTW, I wasn't really drawing an analogy of p as part of a potential,
but simply as something which we might be tempted to regard as
arbitrary, like an additive constant in a potential, if we supposed
that only the divergence of the stress tensor were physical.

With an isotropic pressure you have
div Pi = grad p

I take it that P is the stress tensor?

Hmm... my argument was that in a region of uniform isotropic p (which
corresponds to a divergence-free P), the numerical magnitude of p
still has physical significance (nothing surprising here).

Now, in case the hydrostatic (isotropic) pressure is defined at every
point, but has a non-zero gradient, than the stress tensor is not
divergence-free. That's sensible, since a divergence in the stress
tensor implies a change in momentum density, and a pressure gradient
sounds like it should get things moving. But I wasn't arguing that a
hydrostatic pressure defined at every point implies a divergence free
stress tensor, simply that a _uniform_ hydrostatic pressure, by
itself, is associated with a divergence free stress-tensor.

and with pressure and velocity together,
you also have (in a perfect fluid)

(p/pt) nMu + div (nMuu + p g) = 0

Let's see: in the left hand term you have your intensive rate of
momentum change, so I guess the rest ought to correspond to the
divergence (or minus the divergence?) of the stress tensor.

I would comment at some point that to get identity we must include all
momentum fluxes in the stress tensor (that's what the stress tensor
measures, although this point generates some confusion among
mathematical physicists), but in the Newtonian world, we probably have
to put gravity in by hand: I take it "p" here is yet a third use of
the ascii symbol, as "rho", as well as "partial" and "pressure".

Where nM is the mass density.

Though I may be wrong. :-)

Under gravity it is

(p/pt) nMu + div (nMuu + p g) = -nM grad Phi

At this point I'm losing you. Perhaps nMuu, which looks something
like KE, stands in for a "macroscopic mass flux" part of the momentum
flux, pg for gravitational body force, which leaves the rest to stand
for whatever part of the divergence of the stress tensor (or its
negative) is left over: perhaps reflecting the fact that we are
prejudiced against including the "mass flux" as part of the stress
tensor (aka momentum flux tensor).

where you get Phi from div grad Phi = 4 pi G nM

I'm not sure what Phi or G represent.

In 1-D these things are scalars,

(p/pt) nMu + grad (nMu^2 + p) = -nM grad Phi

and you have a combined potential only in the instance of constant
density,

(p/pt) nMu + grad (nMu^2 + p + nM grad Phi) = 0

The main subtlety is the concept of a total force potential, which
really does have physical reality although it is defined only up to a
constant (which you can easily fold into Phi). You can do this
unambiguously only for constant density.

.

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